The sum of 4th and 6th terms of a geometric series is 80.If the product of the 3rd and 5th term is 256, determine the first term and common ratio
The sum of 4th and 6th terms of a geometric series is 80
---> ar^3 + ar^5 = 80
ar^3(1 + r^2) = 80 or ar^3 = 80/(1+r^2)
the product of the 3rd and 5th term is 256
----> (ar^2)(ar^4) = 256
a^2 r^6 = 256
divide:
ar^3(1 + r^2)/(a^2 r^6) = 80/256
(1+r^2)/ar^3 = 5/16 , but ar^3 = 80/(1+r^2)
(1+ r^2)^2 / 80 = 5/16
(1 + r^2)^2 = 400/16 = 25
1+r^2 = ± 5
r^2 = 4 or r^2 = -6 , the last part is not real
r^2 = 4
r = ± 2
if r = ± 2, sub into a^2 r^6 = 256
64a^2 = 256
a^2 = 4
a = ± 2
so our sequence could be
2, 4, 8, 16, 32, 64, 128, ....
or
2, -4, 8, -16, 32, -64, 128, ..., but that does not satisfy the first condition
or
-2, -4, -8, -16, -32, -64, ... , but that does not satisfy the first condition
or
-2, 4, -8, 16, -32, 64, 128
so a = 2, r = 2
or
a = -2, r = -2
To solve this problem, we can use the formula for the nth term of a geometric series:
an = a1 * r^(n-1)
where:
an is the nth term,
a1 is the first term,
r is the common ratio,
n is the term number.
From the given information, we have two equations:
a4 + a6 = 80 (equation 1)
a3 * a5 = 256 (equation 2)
Let's solve these equations step-by-step:
Step 1: Express a4 and a6 in terms of a1 and r.
Using the formula, we can substitute n = 4 and n = 6:
a4 = a1 * r^(4-1) = a1 * r^3
a6 = a1 * r^(6-1) = a1 * r^5
Step 2: Substitute a4 and a6 into equation 1.
a4 + a6 = 80
a1 * r^3 + a1 * r^5 = 80
Step 3: Simplify the equation.
a1 * r^3 + a1 * r^5 = 80
a1 * (r^3 + r^5) = 80
Step 4: Divide both sides of the equation by a1.
r^3 + r^5 = 80 / a1 (equation 3)
Step 5: Express a3 and a5 in terms of a1 and r.
Using the formula, we can substitute n = 3 and n = 5:
a3 = a1 * r^(3-1) = a1 * r^2
a5 = a1 * r^(5-1) = a1 * r^4
Step 6: Substitute a3 and a5 into equation 2.
a3 * a5 = 256
(a1 * r^2) * (a1 * r^4) = 256
Step 7: Simplify the equation.
(a1 * r^2) * (a1 * r^4) = 256
a1^2 * r^(2+4) = 256
a1^2 * r^6 = 256
Step 8: Take the square root of both sides of the equation.
√(a1^2 * r^6) = √256
Step 9: Simplify the equation.
|a1| * r^3 = 16 (equation 4)
Now we have equations 3 and 4:
r^3 + r^5 = 80 / a1 (equation 3)
|a1| * r^3 = 16 (equation 4)
Since we don't know whether a1 is positive or negative, we can consider two scenarios:
Scenario 1: a1 is positive.
In this case, equation 4 becomes:
a1 * r^3 = 16
We can substitute this into equation 3:
r^3 + r^5 = 80 / (16 / a1) = 5a1
So we have the equation:
r^3 + r^5 = 5a1 (equation 5)
Scenario 2: a1 is negative.
In this case, equation 4 becomes:
-a1 * r^3 = 16
Again, we substitute this into equation 3:
r^3 + r^5 = 80 / (-16 / a1) = -5a1
So we have the equation:
r^3 + r^5 = -5a1 (equation 6)
Now we can solve equations 5 and 6 to find the values of a1 and r.
To find the first term and common ratio of a geometric series, we can use the formulas for the nth term and the sum of a geometric series.
Let's denote the first term as 'a' and the common ratio as 'r'. We can start by finding the 4th and 6th terms.
The formula for the nth term of a geometric series is:
an = a * r^(n-1)
So, the 4th term (a4) would be:
a4 = a * r^(4-1) = a * r^3
Similarly, the 6th term (a6) would be:
a6 = a * r^(6-1) = a * r^5
Now, we are given that the sum of the 4th and 6th terms is 80:
a4 + a6 = 80
a * r^3 + a * r^5 = 80
Next, we are also given that the product of the 3rd and 5th terms is 256:
a * r^(3-1) * a * r^(5-1) = 256
a^2 * r^4 = 256
Now, we have a system of equations with two variables (a and r). We can solve these equations simultaneously to find the values of a and r.
From the first equation, we can factor out 'a' from both terms:
a * (r^3 + r^5) = 80
Dividing both sides by a, we get:
r^3 + r^5 = 80/a ----(Eqn 1)
From the second equation, we can write it as:
(a * r^2) * (a * r^2 * r^2) = 256
(a * r^2)^2 * r^2 = 256
Taking the square root of both sides, we have:
a * r^2 = sqrt(256)
a * r^2 = 16
a * r = 4 ----(Eqn 2)
Now, we can substitute Eqn 2 into Eqn 1 to eliminate 'a':
r^3 + (4 * r) = 80/a
r^3 + 4r = 80/4
r^3 + 4r = 20
r^3 + 4r - 20 = 0
This is a cubic equation in 'r' that we can solve using various methods, such as factoring, using the rational root theorem, or numerical methods like iteration.
Once we find the values of 'r', we can substitute it back into Eqn 2 to find the value of 'a'.
Note: Solving this equation may require advanced mathematical techniques or the use of a calculator or computer algebra system.